help with this question?

320-(74+77+82) or more. Do at least something yourself .

(74+77+82 + x)/4 = 80
233 + x = 320
x = 87

ans x> 87

80 = (74 + 77 + 82 + x) / 4
320 = 233 + x
x = 87

(74 + 77 + 82 + x)/4 ≧ 80
74 + 77 + 82 + x ≧ 320
x ≧ 320 - 74 - 77 - 82
x ≧ 246 - 77 - 82
x ≧ 169 - 82
x ≧ 87

Since the student needs to average 80 on four exams, he/she needs to accumulate 4x80 = 320 points.

Thus far he/she has 74+77+82 = 233 points.

Since 320-233 = 87, 87 additional points are needed.

The score must be x >= 87 which is equivalent to x > 86

From your choices, it looks to me like you should pick x > 87.

You need the average.

(74 + 77 + 82 + x)/4 = 80
(233 + x)/4 = 80
((233 + x)/4) * 4) = 80 * 4
233 + x = 320
233 - 233 + x = 320 - 233
x = 87

The answer is 87.all you have to do is add the numbers 77,74,82 and choose one of the numbers up there and divide by 4(the number of numbers) and you see which one becomes 80!if you add the numbers 77,74,82 you get 233. if you multiply 80 times 4 you get 320 so you subtract 320-233 and you get 87! THAT IS ANOTHER WAY TO DO IT!


im only in 6th!

x>87